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Trignometric equation, proof

  1. Feb 16, 2015 #1

    Suraj M

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    1. The problem statement, all variables and given/known data

    $$ \cos(2x) × \cos(\frac x2) - \cos(3x)× \cos (\frac{9}{2} x) = \sin(5x)× \sin \frac {5}{2}x $$

    2. Relevant equations
    $$ \cos(x) + \cos(y) = 2 \cos{\frac{x+y}{2}} \cos{\frac{x-y}{2}} $$
    $$ \cos(x) - \cos (y) = -2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} $$


    3. The attempt at a solution
    I don't know how to start, as there are no common angles(2x,x/2,3x...).
    How do i start off?
     
    Last edited: Feb 16, 2015
  2. jcsd
  3. Feb 16, 2015 #2

    Orodruin

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    Why do you not simply start expanding your products using the equations you have quoted and see what you get?
     
  4. Feb 16, 2015 #3

    vela

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    First, start by using parentheses when necessary. It's not true that
    $$\cos x + \cos y = 2\cos\left(x+\frac y2\right)\cos\left(x-\frac y2\right)$$ as you wrote.

    It might help to write the identities using variables other than x and y to avoid confusion, e.g.
    $$\cos a + \cos b = 2\cos \frac{a+b}2 \cos \frac {a-b}2.$$ To deal with the term ##\cos 2x \cos \frac x2##, you want ##\frac{a+b}2 = 2x## and ##\frac{a-b}2 = \frac x2##. Solve for ##a## and ##b##. What do you get?
     
  5. Feb 16, 2015 #4

    Suraj M

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    I actually don't know how you right those equations using parentheses, so pardon this:
    ##a = {\frac{5}{2}} x ## and ##b = {\frac{3}{2}} x ##
    and then if i do the same for the second term... ##c = {\frac{15}{2}} x## and ##d = {\frac{-3}{2}} x##
    so then my LHS becomes:
    $${\frac{1}{2}}\left({\cos \left({\frac{5}{2}} x \right) +\cos{\frac{3}{2}}x } \right)- {\frac{1}{2}} \left( [\cos {\frac{15}{2}} x + \cos{\frac{3}{2}} x ] \right) $$
    $$ = {\frac{1}{2}}\left[(-2) × \sin(-5x) × \sin{\frac{5}{2}}x \right] $$
    = RHS
    , Thanks a lot:smile:
    now could you please tell me how to right those equations using parentheses,please
     
    Last edited: Feb 16, 2015
  6. Feb 16, 2015 #5
  7. Feb 16, 2015 #6

    vela

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    cos ((x+y)/2) instead of cos (x+y/2)
     
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